GATE CSE 2024 | Set 1 | Question: 4

Question

Consider a permutation sampled uniformly at random from the set of all permutations of $\{1,2,3, \cdots, n\}$ for some $n \geq 4$. Let $X$ be the event that $1$ occurs before $2$ in the permutation, and $Y$ the event that $3$ occurs before $4$. Which one of the following statements is TRUE?

• A. The events $X$ and $Y$ are mutually exclusive
• B. The events $X$ and $Y$ are independent
• C. Either event $X$ or $Y$ must occur
• D. Event $X$ is more likely than event $Y$

Answer 1

We have to consider random permutation from set {1, 2, 3, - - - - n} for some n $\geq$ 4.

Fast approach to solve this by putting n = 4. 

So, our set will be {1, 2, 3, 4}.

 

Let, X : 1 Occur before 2.

Y : 3 occur before 4.

 

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All Possible Permutations will be :

(1234), (1243), (1324), (1342), (1423), (1432)

(2134), (2143), (2314), (2341), (2413), (2431)

(3124), (3142), (3214), (3241), (3412), (3421)

(4123), (4132), (4213), (4231), (4312), (4321).

 

Now, P(X) = 12/24 = 1/2

P(Y) = 12/24 = 1/2

P(X $\cap$ Y) = 6/24 = 1/4

 

From here, we can observe that P(X $\cap$ Y) = P(X) . P(Y).

So, X and Y are Independent Events.

So, Option B is Correct.

 

Let's not conclude here and try to observe all given options.

 

Option A : X and Y are Mutually Exclusive

X and Y will be Mutually Exclusive iff P(X $\cap$ Y) = 0, but this is not possible here. 

So, Option A is FALSE.

 

Option C : Either Event X or Y must Occur

This statement is FALSE as we have counter-example as : (4213)

So, Option C is FALSE.

 

Option D : Event X is more likely than Event Y.

This statement is clearly FALSE, as P(X) = P(Y).

Hence Event X and Event Y are equally Likely.

So, Option D is FALSE.

 

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Correct Answer : B only

Comments

What is your rank this year?

Answer 2

Just by reading the question, it is evident that -

A. Events X and Y are not mutually exclusive, both events can occur simultaneously.

C. It is also possible that both even can not occur.

D. Both events are symmetric, their probabilities are exactly same.

B. Occurring of either event doesn't affect occurring of another event. Thus, independent.

P(X) = P(Y) = 0.5, P(X,Y) = 0.25 = P(x) * P(Y). This shows that indeed events X and Y are independent.

Total number of permutations = n!

Number of permutations where 1 comes before 2 = ${n \choose 2} * (n-2)!$

Number of permutations where 3 comes before 4 = ${n \choose 2} * (n-2)!$

Number of permutations where 1 comes before 2 and 3 comes before 4 = ${n \choose 2} * {n-2 \choose 2} * (n-4)!$

Answer - B.